1. ## newton's method

Suppose the line y=5x-4 is tangent to the curve y=f(x) when x=3. If newton's method is used to locate a root of the equation f(x)=0 and the initial approximation is x1=3, find the second approximation x2.

can someone help me with this one? thanks

2. Originally Posted by thecount
Suppose the line y=5x-4 is tangent to the curve y=f(x) when x=3. If newton's method is used to locate a root of the equation f(x)=0 and the initial approximation is x1=3, find the second approximation x2.

can someone help me with this one? thanks
Formula:
$\displaystyle x_{n+1} = x_n - f(x_n)/f'(x_n)$

Sovling it:
$\displaystyle x_2 = 3 - f(x)/f'(x) = 3 - 11/5 = 4/5 = 0.8$

The second approximation happens to be the exact answer to the solution. That's because the equation is linear.

3. Originally Posted by TitaniumX
Formula:
$\displaystyle x_{n+1} = x_n - f(x_n)/f'(x_n)$

Sovling it:
$\displaystyle x_2 = 3 - f(x)/f'(x) = 3 - 11/5 = 4/5 = 0.8$

The second approximation happens to be the exact answer to the solution. That's because the equation is linear.
The equation is not linear, at least you are not told that it is, only that $\displaystyle y=5x-4$ is a tangent to $\displaystyle y=f(x)$ at $\displaystyle x=3$. But being a tangent means that you can use it to find the derivative at the point as well as the functional value. Hence that $\displaystyle y=5x-4$ is a tangent allows you to perform one step of Newton's method.

Now you nothing of the true function value at $\displaystyle x=0.8$ nor its derivative so with the given information no further step is possible, but this would only be the exact root if $\displaystyle f(x)=5x-4$, which we are not told.

CB